Einstein's General Theory of Relativity | Lecture 1

Uploaded by StanfordUniversity on 13.01.2009

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Leonard Susskind: Gravity.
Gravity is a rather special force.
It's unusual.
It has difference in electrical forces, magnetic forces, and
it's connected in some way with geometric properties of
space, space and time.
But-- and that connection is, of course, the general theory
of relativity.
Before we start, tonight for the most part we will not be
dealing with the general theory of relativity.
We will be dealing with gravity in its oldest and simplest
mathematical form.
Well, perhaps not the oldest and simplest but Newtonian
And going a little beyond what Newton, certainly nothing
that Newton would not have recognized or couldn't have
grasped-- Newton could grasp anything-- but some ways of
thinking about it which would not be found in Newton's
actual work.
But still Newtonian gravity.
Newtonian gravity is set up in a way that is useful for
going on to the general theory.
Let's, uh, begin with Newton's equations.
The first equation, of course, is F equals MA.
Force is equal to mass times acceleration.
Let's assume that we have a reference, a frame of reference
that means a set of coordinates and that was a set of
clocks, and that frame of reference is what is called an
inertial frame of reference.
An inertial frame of reference simply means one which if
there are no objects around to exert forces on a particular-
- let's call it a test object.
A test object is just some object, a small particle or
anything else, that we use to test out the various fields--
force fields, that might be acting on it.
An inertial frame is one which, when there are no objects
around to exert forces, that object will move with uniform
motion with no acceleration.
That's the idea of an inertial frame of reference.
And so if you're in an inertial frame of reference and you
have a pen and you just let it go, it stays there.
It doesn't move.
If you give it a push, it will move off with uniform
That's the idea of an inertial frame of reference and in an
inertial frame of reference the basic Newtonian equation
number one-- I always forget which law is which.
There's Newton's first law, second law, and third law.
I never can remember which is which.
But they're all pretty much summarized by F equals mass
times acceleration.
This is a vector equation.
I expect people to know what a vector is.
Uh, a three-vector equation.
We'll come later to four-vectors where when space and time
are united into space-time.
But for the moment, space is space, and time is time.
And vector means a thing which is a pointer in a direction
of space, it has a magnitude, and it has components.
So, component by component, the X component of the force is
equal to the mass of the object times the X component of
acceleration, Y component Z component and so forth.
In order to indicate a vector acceleration and so forth I'll
try to remember to put an arrow over vectors.
The mass is not a vector.
The mass is simply a number.
Every particle has a mass, every object has a mass.
And in Newtonian physics the mass is conserved.
It cannot change.
Now, of course, the mass of this cup of coffee here can
It's lighter now but it only changes because mass
transported from one place to another.
So, you can change the mass of an object by whacking off a
piece of it but if you don't change the number of particles,
change the number of molecules and so forth, then the mass
is a conserved, unchanging quantity.
So, that's first equation.
Now, let me write that in another form.
The other form we imagine we have a coordinate system, an X,
a Y, and a Z.
I don't have enough dimensions on the blackboard to draw Z.
It doesn't matter.
X, Y, and Z.
Sometimes we just call them X one, X two, and X three.
I guess I could draw it in.
X three is over here someplace.
X, Y, and Z.
And a particle has a position which means it has a set of
three coordinates.
Sometimes we will summarize the collection of the three
coordinates X one, X two, and X three exactly.
X one, and X two, and X three are components of a vector.
They are components of the position vector of the particle.
The position vector of the particle I will often call either
small r or large R depending on the particular context.
R stands for radius but the radius simply means the distance
between the point and the origin for example.
We're really talking now about a thing with three
components, X, Y, and Z, and it's the radial vector, the
radial vector.
This is the same thing as the components of the vector R.
All right.
The acceleration is a vector that's made up out of a time
derivatives of X, Y, and X, or X one, X two, and X three.
So, for each component-- for each component, one, two, or
three, the acceleration-- which let me indicate, let's just
call it A.
The acceleration is just equal-- the components of it are
equal to the second derivatives of the coordinates with
respect to time.
That's what acceleration is.
The first derivative of position is called velocity.
We can take this thing component by component.
X one, X two, and X three.
The first derivative is velocity.
The second derivative is acceleration.
We can write this in vector notation.
I won't bother but we all know what we mean.
I hope we all know what we mean by acceleration and
And so, Newton's equations are then summarized-- not
summarized but rewritten-- as the force on an object,
whatever it is, component by component, is equal to the mass
times the second derivative of the component of position.
So, that's the summery of-- I think it's Newton's first and
second law.
I can never remember which they are.
Newton's first law, of course, is simply the statement that
if there are no forces then there's no acceleration.
That's Newton's first law.
Equal and opposite.
And so this summarizes both the first and second law.
I never understood why there was a first and second law.
It seemed to me that it was one law, F equals MA.
All right.
Now, let's begin even previous to Newton with Galilean
Gravity how Galileo understood it.
Actually, I'm not sure how much of these mathematics Galileo
did or didn't understand.
Uh, he certainly knew what acceleration was.
He measured it.
I don't know that he had the-- he certainly didn't have
calculus but he knew what acceleration was.
So, what Galileo studied was the motion of objects in the
gravitational field of the earth in the approximation that
the earth is flat.
Now, Galileo knew that the earth wasn't flat but he studied
gravity in the approximation where you never moved very far
from the surface of the earth.
And if you don't move very far from the surface of the
earth, you might as well take the surface of the earth to be
flat and the significance of that is two-fold.
First of all, the direction of gravitational forces is the
same everywheres.
This is not true, of course, if the earth is curved then
gravity will point toward the center.
But the flat space approximation, gravity points down.
Down everywheres always in the same direction.
And second of all, perhaps a little less obvious but
nevertheless true, the approximation where the earth is
infinite and flat, goes on and on forever, infinite and
flat, the gravitational force doesn't depend on how high you
Same gravitational force here as here.
The implication of that is that the acceleration of gravity,
the force apart from the mass of an object, the acceleration
on an object is independent of where you put it.
And so Galileo either did or didn't realize-- again, I don't
know exactly what Galileo did or didn't know.
But what he said was the equivalent of saying that the force
of an object in the flat space approximation is very simple.
It, first of all, has only one component, pointing downward.
If we take the upward sense of things to be positive, then
we would say that the force is-- let's just say that the
component of the force in the X two direction, the vertical
direction, is equal to minus-- the minus simply means that
the force is downward-- and it's proportional to the mass of
the object times a constant called the gravitational
Now, the fact that it's constant everywheres, in other
words, mass times G does vary from place to place.
That's this fact that gravity doesn't depend on where you
are in flat space approximation.
But the fact that the force is proportional to the mass of
an object, that is not obvious.
In fact, for most forces, it is not true.
For electric forces, the force is proportional to the
electric charge, not to the mass.
And so gravitational forces are at a special the strength of
the gravitational force of an object is proportional to its
That characterizes gravity almost completely.
That's the special thing about gravity.
The force is proportional itself to the mass.
Well, if we combine F equals MA with the force law-- this is
the law of force-- then what we find is that mass times
acceleration D second X, now this is the vertical component,
by DT squared is equal to minus-- that is the minus-- MG
That's it.
Now, the interesting thing that happens in gravity is that
the mass cancels out from both sides.
That is what's special about gravity.
The mass cancels out from both sides.
And the consequence of that is that the motion of the
object, its acceleration, doesn't depend on the mass--
doesn't depend on anything about the particle.
The particle, object-- I'll use the word particle.
I don't necessarily mean a point small particle, a baseball
is a particle, an eraser is a particle, a piece of chalk is
a particle.
That the motion of an object doesn't depend on the mass of
the object or anything else.
The result of that is if you take two objects of quite
different mass and you drop them, they fall exactly the same
Galileo did that experiment.
I don't know whether he really threw something off the
Leaning Tower of Pisa or not.
It's not important.
He did balls down an inclined plane.
I don't know whether he actually did or didn't.
I know the myth is that he didn't.
I find it very difficult to believe that he didn't.
I've been in Pisa.
Last week I was in Pisa and I took a look at the Leaning
Tower of Pisa.
Galileo was born and lived in Pisa.
He was interested in gravity.
How it would be possible that he wouldn't think of dropping
something off the Leaning Tower is beyond my comprehension.
You look at that tower and say, "That tower is good for one
thing: Dropping things off. "
[laughing] Leonard Susskind: Now, I don't know.
Maybe the doge or whoever they called the guy at the time
said, no, no Galileo.
You can't drop things from the tower.
You'll kill somebody.
So, maybe he didn't.
He must have surely thought of it.
All right.
So, the result, had he done it, and had he not had to worry
about such spurious effects as air resistance would be that
a cannon ball and a feather would fall in exactly the same
way, independent of the mass, and the equation would just
say, the acceleration would first of all be downward, that's
the minus sign, and equal to this constant G.
Excuse me.
Now, G as a number, it's 10 meters per second per second at
the surface of the earth.
At the surface of the moon it's something smaller.
On the surface of Jupiter it's something larger.
So, it does depend on the mass of the planet but the
acceleration doesn't depend on the mass of the object you're
It depends on the mass of the object you're dropping it onto
but not the mass of the object stopping it.
That fact, that gravitational motion, is completely
independent of mass is called or it's the simplest version
of something that's called the equivalence principle.
Why it's called the equivalence principle we'll come to
What's equivalent to what.
At this stage we can just say gravity is equivalent between
all different objects independent of their mass.
But that is not exactly what is equivalence/inequivalence
principle was all about.
All right.
That has a consequence.
An interesting consequence.
Supposing I take some object which is made up out of
something which is very unrigid.
Just a collection of point masses.
Maybe let's even say they're not even exerting any forces on
each other.
It's a cloud, a very diffuse cloud of particles and we watch
it fall.
Now, let's suppose we start each particle from rest, not all
at the same height, and we let them all fall.
Some particles are heavy, some particles are light, some of
them may be big, some of them may be small.
How does the whole thing fall? And the answer is, all of the
particles fall at exactly the same rate.
The consequence of it is that the shape of this object
doesn't deform as it falls.
It stays absolutely unchanged.
The relationship between the neighboring parts are
There are no stresses or strains which tend to deform the
So even if the object were held together but some sort of
struts or whatever, there would be no forces on those struts
because everything falls together.
Okay? The consequence of that is that falling in the
gravitational field is undetectable.
You can't tell that you're falling in a gravitational field
by-- when I say you can't tell, certainly you can tell the
difference between free fall and standing on the earth.
All right? That's not the point.
The point is that you can't tell by looking at your
neighbors or anything else that there's a force being
exerted on you and that that force that's being exerted on
you is pulling downward.
You might as well, for all practical purposes, be infinitely
far from the earth with no gravity at all and just sitting
there because as far as you can tell there's no tendency for
the gravitational field to deform this object or anything
You cannot tell the difference between being in free space
infinitely far from anything with no forces and falling
freely in a gravitational field.
That's another statement of the equivalence principle.
>> You say not mechanically detectable? Leonard Susskind:
Well, in fact, not detectable, period.
But so far not mechanically detectable.
>> Well, would it be optically detectable? Leonard Susskind:
For example, these particles could be equipped with lasers.
Lasers and optical detectors of some sort.
What's that? Oh, you could certainly tell if you were
standing on the floor here you could definitely tell that
there was something falling toward you.
But the question is, from within this object by itself,
without looking at the floor, without knowing that the floor
was-- >> Something that wasn't moving.
Leonard Susskind: Well, you can't tell whether you're
falling and it's, uh-- yeah.
If there was something that was not falling it would only be
because there was some other force on it like a beam or a
tower of some sort holding it up.
Why? Because this object, if there are no other forces on
it, only the gravitational forces, it will fall at the same
rate as this.
All right.
So, that's another expression of the equivalence principle,
that you cannot tell the difference between being in free
space far from any gravitating object versus being in a
gravitational field.
Now, we're gonna modify this.
This, of course, is not quite true in a real gravitational
field, but in this flat space approximation where everything
pulls together, you cannot tell that there's a gravitational
At least, you cannot tell the difference-- not without
seeing the floor in any case.
The self-contained object here does not experience anything
different than it would experience far from any gravitating
object standing still.
Or in uniform motion.
>> Another question.
Leonard Susskind: What's that? Yeah.
>> We can tell where we're accelerating.
Leonard Susskind: No, you can't tell when you're
>> Well, you can-- you can't feel-- isn't that because you
can tell there's no connection between objects? Leonard
Susskind: Okay.
Here's what you can tell.
If you go up to the top of a high building and you close
your eyes, and you step off, and go into free fall, you will
feel exactly the same-- you will feel weird.
I mean, that's not the way you usually feel because your
stomach will come up and do some funny things.
You know, you might lose it.
But the point is, you would feel exactly the same discomfort
in outer space far from any gravitating object just standing
You'll feel exactly the same peculiar feelings.
Okay? What are those peculiar feelings due to? They're not
due to falling.
They're due to not fall-- well. . .
[laughing] they're due to the fact that when you stand on
the earth here, there are forces on the bottom of your feet
which keep you from falling and if the earth suddenly
disappeared from under my feet, sure enough, my feet would
feel funny because they're used to having that force exerted
on their bottoms.
You get it.
I hope.
So, the fact that you feel funny in free fall, uh, is
because you're not used to free fall.
It doesn't matter whether you're infinitely far from any
gravitating objects standing still or freely falling in the
presence of a gravitational field.
Now, as I said, this will have to be modified in a little
There are such things as tidal forces.
Those tidal forces are due to the fact that the earth is
curved and that the gravitational field is not the same in--
the same direction in every point, and that it varies with
That's due to the finiteness of the earth.
But, in the flat space of the-- in the flat earth
approximation where the earth is infinitely big pulling
uniformly, uh, there is no other effect of gravity that is
any different than being in free space.
Again, that's known as the equivalent principle.
Now, let's go on beyond the flat space or the flat earth
approximation and move on to Newton's theory of gravity.
Newton's theory of gravity says every object in the universe
exerts a gravitational force on every other object in the
Let's start with just two of them.
Equal and opposite.
Attractive means that the direction of the force on one
object is toward the other one.
Equal and opposite forces and the magnitude of the force--
the magnitude of the force of one object on another.
Let's characterize them by a mass.
Let's call this one little m.
Think of it as a lighter mass and this one, which we can
imagine as a heavier object, we'll call it big M.
All right.
Newton's law of force is that the force is proportional to
the product of the masses.
Making either mass heavier will increase the force.
The product of the masses, big M times little m, inversely
proportional to the square of the distance between them.
Let's call that R squared.
Let's call the distance between them R.
And there's a numerical constant.
This law by itself could not possibly be right.
It's not dimensionally consistent.
The-- if you work out the dimensions of force, mass, mass
and R, it's not dimensionally consistent.
There has to be a constant in there.
And that numerical constant is called capital G, Newton's
And it's very small.
It's a very small constant.
I'll write down what it is.
G is equal to six or 6. 7, roughly, times ten to the minus
11th, which is a small number.
So, on the face of it, it seems that gravity is a very weak
Umm, you might not think that gravity is such a weak force,
but to convince yourself it's a weak force there's an
experiment that you can do.
Weak in comparison to other forces.
I've done this for classes and you can do it yourself.
Just take an object hanging by a string and two experiments.
The first experiment, take a little object here and
electrically charge it.
You electrically charge it by rubbing it on your shirt.
That doesn't put much charge on it but it charges it up
enough to feel some electrostatic force.
Then take another object of exactly the same kind, rub it on
your shirt, and put it over here.
What happens? They repel.
And the fact that they repel means that this will shift.
And you'll see it shift.
Take another example.
Take your little ball there to be iron and put a magnet next
to it.
Again, you'll see quite an easily detectable deflection of
the-- of the string holding it.
All right? Next, take a 10,000-pound weight and put it over
Guess what happens? Undetectable.
You cannot see anything happen.
The gravitational force is much, much weaker than most other
kinds of forces and that's due to-- not due to.
Not due to that.
The fact that it's so weak is encapsulated in this small
number here.
Another way to say it is if you take two masses, each of 1
kilometer-- not 1 kilometer.
1 kilogram.
A kilogram is a good healthy mass, a nice chunk of iron.
MM and you separate them from 1 meter, then the force
between them is just G and it's 6. 7 times ten to the minus
11th, if you do it with the units being Newton's.
Very weak force.
But, weak as it is, we feel it rather strenuously.
We feel it strongly because the earth is so darn heavy.
So, the heaviness of the earth makes up for the smallness of
G and so we wake up in the morning feeling like we don't
wanna get out of bed because gravity is holding us down.
[laughing] Leonard Susskind: Yeah? >> Umm, so that force is
measuring the force between-- from the large one to the
small one or both? Leonard Susskind: Both.
They're equal and opposite.
Equal and opposite.
That's the rule.
That's Newton's third law.
The forces are equal and opposite.
So, the force on the large one due to the small one is the
same as the force of the small one on the large one.
But it is proportional to the product of the masses.
So, the meaning of that is I'm not-- I'm heavier than I'd
like to be but I'm not very heavy.
I'm certainly not heavy enough to deflect the hanging weight
But I do exert a force on the earth which is exactly equal
and opposite to the force that the very heavy earth exerts
on me.
Why does the earth accel-- if I drop from a certain height,
I accelerate down.
The earth hardly accelerates at all, even though the forces
are equal.
Why is it that the earth-- if the forces are equal, my force
on the earth and the earth's force on me are equal, why is
it that the earth accelerates so little and I accelerate so
much? Yeah.
Because the acceleration involves two things.
It involves the force and the mass.
The bigger the mass, the less the acceleration for the
So, the earth doesn't accelerate-- yeah, question.
>> How did Newton arrive at that equation for the
gravitational force? Leonard Susskind: I think it was
largely a guess.
But it was an educated guess.
And, umm, what was the key-- it was large-- no, no.
It was from Kepler's laws.
It was from Kepler's laws.
He worked out, roughly speaking-- I don't know what he did.
He was secretive and he didn't really tell people what he
But, umm, the piece of knowledge that he had was Kepler's
laws of motion-- planetary motion-- and my guess is that he
just wrote down a general force and realized that he would
get Kepler's laws of motion for the inverse square law.
Umm, I don't believe he had any understand lying theoretical
reason to believe in the inverse square law.
>> Edmund Halley actually asked him, uh, what kind of force
law do you need for conic section orbits and he had almost
performed the calculations a year before.
Leonard Susskind: Yeah.
>> So, yeah.
Leonard Susskind: Actually, I don't think-- yeah.
>> I think the question-- he asked the question for inverse
square laws and I think that Newton already knew the
solution was an ellipse.
Leonard Susskind: No.
It wasn't the ellipse that was there.
The orbits might have been circular.
It was the fact that the period varies as the three halfs
power of the radius.
All right? The period of motion is the circular motion has
an acceleration toward the center.
Any motion of the circle is accelerated toward the center.
If you know the period and the radius, then you know the
acceleration toward the center.
Okay? We could write the-- what's the word? Anybody know
what-- if I know the angular frequency of going around in an
orbit that's called omega.
Anybody know the-- and it's basically just the inverse
Okay? Omega is roughly the inverse period number of cycles
per second.
What is the acceleration of a thing moving in a circular
Anybody remember? >> Omega squared R.
Leonard Susskind: Omega squared R.
That's the acceleration.
Now, supposing he sets that equal to some unknown force law
F of R and then divides by R.
Then he finds omega as a function of the radius of the
Well, let's do it for the real case.
For the real case, inverse square law, F of R is one of R
squared so this would be one of R cubed and in that form it
is Kepler's second law? I don't even remember which one it
It's the law that says that the frequency or the period, the
square of the period, is proportional to the cube of the
That was the law of Kepler.
So, from Kepler's laws he easily could-- or that one law, he
could easily reduce that the force was proportional to one
of R squared.
I think that's probably historically what he did.
Then, on top of that he realized that if you department have
a perfectly circular law orbit then the inverse square law
was the unique law which would give elliptical orbits.
So, it's a two-step thing.
>> What happens when the two objects are touching? Do you
measure it from the-- Leonard Susskind: Of course, there are
other forces on them.
If two objects are touching, there are all sorts of forces
between them that are not just gravitational.
Electrostatic forces, atomic forces, nuclear forces? So,
you'll have to modify the whole story.
>> As the distance approaches zero-- Leonard Susskind: Yeah.
Then it breaks down.
Then it breaks down.
Then it breaks down.
When they get so close that other important forces come into
The other important forces, for example, are the forces that
are holding this object and preventing it from falling.
These we usually call them contact forces but, in fact, what
they really are is various kinds of electrostatic forces
between the atoms and molecules on the table and the atoms
and molecules in here.
So, other kinds of forces.
All right.
Incidentally, let me just point out if we're talking about
other kinds of force laws, for example, electrostatic force
laws, then the force-- we still have F equals MA but the
force law-- the force law will not be that the force is
somehow proportional to the mass times something else but it
could be the electric charge.
If it's the electric charge, then electrically uncharged
objects will have no forces on them and they won't
Electrically charged objects will accelerate in an electric
So, electrical forces don't have this universal property
that everything falls or everything moves in the same way.
Uncharged particles move differently than charged particles
with respect to electrostatic forces.
They move the same way with respect to gravitational forces.
And as repulsion and attraction, whereas gravitational
forces are always attractive.
Where is my gravitational force? I lost it.
Here it is.
All right.
So, that's Newtonian gravity between two objects.
For simplicity let's just put one of them, the heavy one, at
the origin of coordinates and study the motion of the light
one then-- oh, incidentally, one usually puts-- let me
refine this a little bit.
As I've written it here, I haven't really expressed it as a
vector equation.
This is the magnitude of the force between two objects.
Thought of as a vector equation, we have to provide a
direction for the force.
Vectors have directions.
What direction is the force on this particle? Well, the
answer is, it's along the radial direction itself.
So, let's call the radial distance R, or the radial vector
R, then the force on little m here is along the direction R.
But it's also opposite to the direction of R.
The radial vector, relative to the origin over here, points
this way.
On the other hand, the force points in the opposite
If we wanna make a real vector equation out of this, we
first of all have to put a minus sign.
That indicates that the force is opposite to the direction
of the radial distance here, but we also have to put
something in which tells us what direction the force is in.
It's along the radial direction.
But wait a minute.
If I multiply it by R up here, I had better divide it by
another factor of R downstairs to keep the magnitude
The magnitude of the force is one over R squared.
If I were to just randomly come and multiply it by R, that
would make the magnitude bigger by a factor of R, so I have
to divide it by the magnitude of R.
This is Newton's force law expressed in vector form.
Now, let's imagine that we have a whole assembly of
A whole bunch of them.
They're all exerting forces on one another.
In pairs, they exert exactly the force that Newton wrote
But what's the total force on a particle? Let's label these
particles the first one, the second one, the third one, the
fourth one, dot, dot, dot, dot, dot.
This is the Ith one over here.
So, I is the running index which labels which particle we're
talking about.
The force on the Ith particle, let's call it F sub I, and
let's remember that it's a vector, it's equal to the sum--
now, this is not an obvious fact that when you have two
objects exerting a force on the third that the force is
necessarily equal to the sum of the two forces, of the two
You know what I mean.
But it is a fact anyway.
Obvious or not obvious it is a fact.
Gravity does work that way at least in the Newtonian
With Einstein, it breaks down a little bit.
But in Newtonian physics the force is the sum and so it's a
sum of all the other particles.
Let's write that J not equal to I.
That means it's a sum over all not equal to I.
So, the force from the first particle.
It comes from the second particle, third particle, third
particle, and so forth.
Each individual force involves M sub I, the force of the Ith
particle, times the four, times the mass of the Jth
Product of the masses divided by the square of the distance
between them, let's call that RIJ squared.
The distance between the Ith particle is I and J, the
distance between the Ith particle and the Jth particle is
But then, just as we did before, we have to give it a
Put a minus sign here, that indicates that it's attractive,
another RIJ upstairs, but that's a vector RIJ, and make this
cubed downstairs.
All right? So, that says that the force on the Ith particle
is the sum of all the forces due to all the other ones of
the product of their masses inverse square in the
denominator, and the direction of each individual force on
this particle is toward the other.
All right? This is a vector sum.
Yeah? Hmm? The minus indicates that it's attractive.
>> but you've got the vector going from I to J.
Leonard Susskind: Oh.
Let's see.
That's the vector going from R to I to J.
There is a question of the sign of this vector over here.
So yeah.
You're absolute-- yeah.
I actually think it's-- yeah, you're right.
You're absolutely right.
The way I've written it there should not be a minus sign
If I put RJI there, then there would be a minus sign.
Right? So, you're right.
But in any case every one of the forces is attractive and
what we have to do is to add them up.
We have to add them up as vectors and so there's some
resulting vector, some resultant vector, which doesn't point
toward any one of them in particular but points in some
direction which is determined by the vector sum of all the
All right in but the interesting fact is, if we combine
this, this is the force on the Ith particle.
If we combine it with Newton's equations-- let's combine it
with Newton's F equals MA equations then this is F.
This on the Ith particle, this is equal to the mass of the
Ith particle times the acceleration of the Ith particle.
Again, vector equations.
Now, the sum here is over all the other particles.
We're focusing on number I.
I, the mass of the Ith particle will cancel out of this
I don't wanna throw it away but let's just circle it and now
We notice that the acceleration of the Ith particle does not
depend on its mass again.
Once again, because the mass occurs on both sides of the
equation it can be canceled out, and the motion of the Ith
particle does not depend on the mass of the Ith particle.
It depends on the masses of all the other ones.
All the other ones come in, but the mass of the Ith particle
cancels the equation.
So, what that means is if we had a whole bunch of particles
here and we added one more over here, its motion would not
depend on the mass of that particle.
It depends on the mass of all the other ones but it doesn't
depend on the mass of the Ith particle here.
Okay? Again, equivalence principle that the motion of a
particle doesn't depend on its mass.
And again if we had a whole bunch of particles here, if they
were close enough together, they would all move in the same
Before I discuss any more mathematics, let's just discuss
tidal forces, what tidal forces are.
>> Can I ask one question? Leonard Susskind: Yeah.
>> Once you set this whole thing into motion dynamically.
Leonard Susskind: Yeah.
>> We have all different masses and each particle is gonna
be effected by each one? Leonard Susskind: Yes.
>> Every particle in there is going to experience a uniform
acceleration? Leonard Susskind: No, no, no.
The acceleration is not uniform.
The acceleration will get larger when it gets closer to one
of the particles.
It won't be uniform anymore.
It won't be uniform now because the force is not independent
of where you are.
Now the force depends on where you are relative to the
objects that are exerting the force.
It was only in the flat earth approximation where the force
didn't depend on where you were.
Okay? Now, the force varies so it's larger where you're far
away-- sorry.
It's smaller when you're far away, it's smaller when you're
in close.
Okay? >> But is it going to be changing in a-- it changes in
a vector form with each individual particle.
Each one of them is changing position.
Leonard Susskind: Yeah.
>> So, is the dynamics that every one of them is going
towards the center of gravity of the entire-- Leonard
Susskind: Not necessarily.
I mean, they could be flying apart from each other but they
will be accelerating toward each other.
Okay? If I throw this eraser in the air with greater than
the escape velocity, it's not going to turn around and fall
back down.
>> Well, the question is, is the acceleration a uniform
acceleration or is it changing in dynamics? Leonard
Susskind: Changing with what? With respect to what? Time?
It changes with respect to time because the object moves
farther and farther away.
>> In the two-mass system-- Leonard Susskind: Mm hmm.
>> I call that a uniform acceleration.
Leonard Susskind: Uniform with respect to what? >> It's not
The radius is changing and it's inversed cubically radiused.
Leonard Susskind: Inverse squared.
>> Inverse squared.
Leonard Susskind: Let's take the earth.
Here's the earth, and we drop a small mass from far away.
As that mass moves in, its acceleration increases.
Why does its acceleration increase in its acceleration
increases because the radial distance gets smaller.
So, in that sense it's not.
All right.
Now, once the gravitational force depends on distance then
it's not really quite true that you don't feel anything in a
gravitational field.
You feel something that's to some extent different than you
would feel in free space without any gravitational field.
The reason is more or less obvious.
Here you are-- here's the earth.
Now, you, or me, or whoever it is, happens to be extremely
Couple of thousand miles tall.
Well, this person's feet are being pulled by the
gravitational field more than his head.
Or another way of saying the same thing is if let's imagine
that the person is very loosely held together.
[laughing] Leonard Susskind: He's more or less a gas of-- we
are pretty loosely held together.
At least I am.
All right.
The acceleration on the lower portions of his body are
larger than the accelerations on the upper part of his body.
So it's quite clear what happens to him.
You get stretched.
He doesn't get a sense of falling as such.
He gets a sense of stretching, being stretched.
Feet being pulled away from his head.
At the same time, let's-- all right.
So let's change his shape a little bit.
I just spent a week-- two weeks in Italy and my shape
changes whenever I go to Italy.
It tends to get more horizontal.
My head is here, my feet are here, and now I'm this way.
Still loosely put together.
All right? Now what? Well, not only does the force depend on
the distance but it also depends on the direction.
The force on my left end over here is this way.
The force on my right end over here is this way.
The force on the top of my head is down but it's weaker than
the force on my feet.
So there are two effects.
One effect is to stretch me vertically.
It's because my head is not being pulled as hard as my feet.
But the other effect is to be squished horizontally by the
fact that the forces on the left end of me are pointing
slightly to the right and the forces to the right end of me
are pointing slightly to the left.
So a loosely knit person like this falling in free fall near
a real planet or a real gravitational object which has a
real Newtonian gravitational field around it will experience
a distortion-- will experience a degree of distortion and a
feeling of being stretched vertically, being compressed
horizontally, but if the object is small enough-- what does
small enough mean? Let's suppose the object that's falling
is small enough.
If it's small enough, then the gradient of the gravitational
field across the size of the object will be negligible and
so all parts of it will experience the same gravitational
All right.
So tidal forces-- these are tidal forces.
These forces which tend to tear things apart vertically and
squish them this way.
Tidal forces.
Tidal forces are forces which are real.
You feel them.
I mean-- yeah? >> Do you recall if Newton calculated lunar
tides? Leonard Susskind: Oh, I think he did.
He certainly knew the cause of the tides.
I don't know to what extent he calculated.
What do you mean calculated the-- >> As in this kind of a
system with the moon and the sun.
Leonard Susskind: Well, I doubt that he was capable-- I'm
not sure whether he estimated the height of the deformation
of the oceans or not.
But I think he did understand this much about tides.
So, that's what's called tidal force.
And remember, tidal force has this effect of stretching and
in particular if we take the earth -- just to tell you why
it's called tidal forces of course is because it has to do
with tides.
I'm sure you all know the story.
But if this is the moon down here, then the moon exerting
forces on the earth exerts tidal forces on the earth, which
means to some extent it tends to stretch it this way and
squash it this way.
Well, the earth is pretty rigid so it doesn't form very much
due to the moon.
But what's not rigid is the layer of water around it and so
the layer of water tends to get stretched and squeezed and
so it gets deformed into a deformed shell of water with a
bump on this side and a bump on that side.
All right.
I'm not gonna go any more deeply into that than I'm sure
you've all seen.
But let's define now what we mean by the gravitational
The gravitational field is abstracted from this formula.
We have a bunch of particles-- >> question.
Leonard Susskind: Yeah? >> Don't you have to use some sort
of coordinate geometry so that when you have the poor guy in
the middle's being pulled by all the other guys on the side.
Leonard Susskind: I'm not explaining it right.
>> it's always negative, is that what you're saying? Leonard
Susskind: No.
It's always attractive.
All right.
So you have-- >> What about the other guys that are pulling
upon him from different directions? Leonard Susskind: Let's
suppose it's somebody over here and we're talking about the
force on this person over here.
Obviously there's one force pushing this way and another
force pushing that way.
Okay? No.
They're all opposite to the direction of the object which is
pulling on him.
All right? That's how this mind of science says.
>> well, you kind of retracted the minus sign at the front
and reversed the JI.
So it's the direction-- Leonard Susskind: We can get rid of
the minus sign in the front there by switching this RJ.
RIJ and RJI are opposite to each other.
One of them is the vector between I and J.
I and J.
And the other is the vector to J and I, so they're equal and
opposite to each other.
The minus sign there.
Look, as far as the minus sign goes, all that it means is
every one of these particles is pulling on this particle
toward it as opposed to pushing away from it.
It's just a convention which keeps track of attraction
instead of repulsion.
>> yeah.
For the Ith mass-- if that's the right word.
Leonard Susskind: Yeah.
>> If you look at it as kind of an ensemble wouldn't there
be a nonlinear component to it was the I guy, the Ith guy,
the Jth guy, then with you compute the Jth guy-- you know
what I mean? Leonard Susskind: When you take into account
the motion.
Now, what this formula is for is supposing you know the
positions of all the others.
You know that.
All right? Then what is the force on one additional one in
but you're perfectly right.
Once you let the system evolve, then each one will cause a
change in motion of the other one and so it because a
complicated as you say nonlinear mess.
But this formula is a formula for if you knew the position
and location of every particle, this would be the force.
Okay? Something.
You need to solve the equations to know how the particles
But if you know where they are, then this is the force on
the Ith particle.
All right.
Let's come to the idea of the gravitational field.
The gravitational field is in some ways similar to the
electric field of an electric charge.
It's the combined effect of all the masses everywheres.
And the way you define it is as follows: You imagine one
more particle, one more particle.
You can take it to be a very light particle so it doesn't
influence the motion of the others.
Add one more particle.
In your imagination.
You don't really have to add it.
In your imagination.
And what the force on it is.
The force is the sum of the force due to all the others.
It is proportional.
Each term is proportional to the mass of this extra
This extra particle which may be imaginary is called a test
It's a thing that you're imagining testing out the
gravitational field with.
You take a light little particle, and you put it here, and
you see how it accelerates.
Knowing how it accelerates tells you how much force is on
In fact, it just tells you how it accelerates.
And you can go around and imagine putting it in different
places and mapping out the force field that's on that
Or the acceleration field since we already know that the
force is proportional to the mass.
Then we can just concentrate on the acceleration.
The acceleration all particles will have the same
acceleration independent of the mass.
So we don't even have to know what the mass of the particle
We put something over there, a little bit of dust, and we
see how it accelerates.
Acceleration is a vector and so we map out in space the
acceleration of a particle at every point in space, either
imaginary or real particle, and that gives us a vector field
at every point in space.
Every point in space there is a gravitational field of
It can be thought of as the acceleration.
You don't have to think of it as force.
The acceleration of a point mass located at that position.
It's a vector that has a direction, it has a magnitude, and
it's a function of position.
So, we just give it a name.
The acceleration due to all the gravitating objects is a
vector and it depends on position.
Your X means location.
It means all of the components of position X, Y, and Z, and
it depends on all the other masses in the problem.
That is what's called the gravitational field.
It's very similar to the electric field except the electric
field is force per unit charge.
It's the force of an object divided by the charge on the
The gravitational field is the force on the object divided
by the mass on the object.
Since the force is proportional to the mass the acceleration
field did you want depend on which kind of particle we're
talking about.
All right.
So, that's the idea of a gravitational field.
It's a vector field and it varies from place to place.
And, of course, if the particles are moving, it also varies
in time.
If everything is in motion, the gravitational field will
also depend on time.
We can even work out what it is.
We know what the force on the Ith particle is.
Right? The force on a particle is the mass times the
So, if we wanna find the acceleration, let's take the Ith
particle to be the test particle.
Little i represents the test particle over here.
Let's erase the immediate step over here and write that this
is MI times AI but let me call it now capital A.
The acceleration of a particle at position X is given by the
right-hand side.
And we can cross out the MI because it cancels from both
So, here's a formula for the gravitational field at an
arbitrary point due to a whole bunch of massive objects.
A whole bunch of massive objects.
An arbitrary particle put over here will accelerate in some
direction that's determined by all the others and that
acceleration is gravitation-- definition.
Definition is the gravitational field.
Let's take a little break.
We usually take a break at about this time and I recover my
To go on, we need a little bit of fancy mathematics.
We need a piece of mathematics called Gauss's theorem and
Gauss's theorem involves integrals, derivatives,
And we need to spell those things out.
They're essential part of the theory of gravity.
And much of these things that we've done in the context of
the electrical forces, in particular the concept of
divergence, divergence of a vector field.
So, I'm not going to spend a lot of time on it.
If you need to fill in, then I suggest you just find any
book on vector calculus and find out what a divergence, and
a gradient, and a curl-- we won't do curl today.
What those concepts are, and look up Gauss's theorem and
they're not terribly hard but we're gonna go through them
fairly quickly here since we've done them several times in
the past.
All right.
Imagine that we have a vector field.
Let's call that vector field A.
It could be the field of acceleration and that's the way I'm
gonna use it.
But for the moment it's just an arbitrary vector field, A.
It dependence on position.
When I say it's a field, the implication is that it depends
on position.
Now I probably made it completely unreadable.
A of X varies from point to point.
And I want to define a concept called the divergence of a
Now, it's called a divergence because what it has to do is
the way the field is spreading out away from the point.
For example, a characteristic situation where we would have
a strong divergence for a field is if the field was
spreading out from a point, like that.
The field is diverging away from the point.
Incidentally, after the field is pointing inward, then one
might say the field has a convergence but we simply say it
has a negative divergence.
So, divergence can be positive or negative.
And there's a mathematical expression which represents the
degree to which the field is spreading out like that.
It is called the divergence.
I'm gonna write it down and it's a good thing to get
familiar with, certainly if you're going to follow this
course it's a good thing to get familiar with.
But if you're gonna follow any kind of physics course past
freshmen physics, the idea of divergence is very point.
All right.
Supposing the field A has a set of components.
The one, two, and three component or we could call them the
X, Y, and Z component.
Now I'll use X, Y, and Z.
X, Y, and Z.
Which I previously called X one, X two, and X three.
It has components at AX, AY, and AZ.
Those are the three components of the vector.
Well, the divergence has to do, among other things, with the
way the field varies in space.
If the field is the same everywheres in space what would
that mean? That would men that the field has not only the
same magnitude, but the same direction anywheres in space.
Then it just points in the same direction everywheres in
space with the same magnitude.
It certainly has no tendency to spread out.
When does a field have a tendency to spread out in when a
field varies.
For example, it could be small over here, growing bigger,
growing bigger, growing bigger.
And we might even go in the opposite direction and discover
that it's the opposite direction getting bigger in that
Now, clearly there's a tendency for the field to spread out
from the center here.
The same thing could be true if it were varying in the
vertical direction or if it was varying in the other
horizontal direction.
And so the divergence, whatever it is, has to do with
derivatives of the components of the field.
I'll just tell you exactly what it is.
It is equal to it.
The divergence of a field is written this way: Upside down
The meaning of this symbol, the meaning of an upside down
triangle is always that it has to do with derivatives, the
three derivatives.
Derivatives, whether it's the three partial derivatives.
Derivative with respect to X, Y, and Z.
And this is by definition.
The derivative with respect to X of the X component of A
plus the derivative with respect to Y of the Y component of
A, plus the derivative with respect to Z of the Z component
of A.
That's definition.
What's not a definition is the theorem and it's called
Gauss's theorem.
>> I'm sorry.
Is that a vector or is it-- Leonard Susskind: No.
That's a scale of quantity.
It's a scale of quantity.
It's a scale of quantity.
So, let me write it.
It's the derivative of A sub X with respect to X, that's
what this means, plus the derivative of ace of Y with
respect of Y, plus the derivative of ace of Z with respect
to Z.
>> Yeah.
So, the arrows you were drawing over there they were just A
on the other board.
You drew some arrows on the other board that are now hidden.
Leonard Susskind: Yeah.
>> Those were just A? Leonard Susskind: Yeah.
>> Not the divergence.
Leonard Susskind: Right.
Those were A.
And A has a divergence when it's spreading away from a
point, but a divergence is itself a scale of quantity.
Let me try to give you some idea of what divergence means in
a context where you can visualize it.
Imagine that we have a flat lake.
Just a shallow lake.
And the water is coming up from underneath.
It's being pumped in from somewheres underneath.
What happens if the water's being pumped in.
Of course, it tends to spread out.
Let's assume that depth can't change.
We put a lid over the whole thing so that it can't change
its depth.
We pump some water in from underneath and it spreads out.
Okay? We suck some water out from underneath and it spreads
It anti-spreads.
So, the spreading water has a divergence.
Water coming in towards the place where it's being sucked
out it has a convergence or a negative divergence.
Now, we can be more precise about that.
We look down at the lake from above, and we see all the
water is moving of course.
If it's being pumped in the water is moving.
And there is a velocity vector.
At every point there is a velocity vector.
So, at every point in this lake there's a velocity vector
and in particular if there's water being pumped in from the
center here, right? Underneath the water of the lake there's
some water being pumped in the water's being sucked in from
that point.
Okay? And there'll be a divergence where the water is being
pumped in.
Okay if the water is being pumped out then exactly the
The arrows point inward and there's a negative divergence.
If there's no divergence, then, for example, a simple
situation with no divergence.
That doesn't mean the water is not moving.
But a simple example of no divergence is the water is all
moving together.
You know, the river is simultaneous, the lake is moving
simultaneously in the same direction with the same velocity.
It can do that without any water being pumped in.
But if you found that the water was moving from the right on
this side and the left on that side, you'd be pretty sure
that somewheres in between, water had to be pumped in.
Right? If you found the water was spreading out away from a
line this way here and this way here, then you'd be pretty
sure that some water was being pumped in from underneath
along this line here.
Well, you would see in another way you would discover that
the X component of the velocity has a derivative.
It's different over here than it is from over here.
The X component of the velocity varies along the X
So, the fact that the X component of the velocity is varying
along the X direction is an indication that there's some
water being pumped in here.
Likewise, if you discovered that the water was flowing up
over here and down over here, you would expect it in here
somewheres some water was being pumped in.
So, derivatives of the velocity are often an indication that
there's some water being pumped in from underneath.
That pumping in of the water is the divergence of the
velocity vector.
Now, the water, of course, is being pumped in from
So, there's a direction of flow but it's coming from
There's no sense of direction-- well, okay.
That's what divergence is.
>> I have a question.
The diagrams you already have on the other board behind you?
Leonard Susskind: Yeah.
>> With the arrows? Leonard Susskind: Yeah.
Leonard Susskind: If you take, say, the right-most arrow and
you draw a circle between the head and tail in between, then
you can see the in and the out.
Leonard Susskind: Mm hmm.
>> The in arrow and the out arrow of a certain right in
between those two.
And let's say that the bigger arrow's created by a steeper
slope of the streak.
Leonard Susskind: No, this is faster.
>> It's going faster.
>> Okay.
And because of that, there's a divergence there that's
basically it's sort of the difference between the in and the
Leonard Susskind: That's right.
That's right.
If we draw a circle around here or we would see that-- the
water is moving faster over here than it is over here, more
water is flowing out over here than is coming in over here.
Where's it coming from? It must be coming in.
The fact that there's more water flowing out on one side
than is coming in from the other side must indicate that
there's a net inflow from somewheres else and the somewheres
else would be from the pumping water from underneath.
So, that's the idea of divergence.
>> Could it also be because it's thinning out? Could that be
a crazy example? Like, the lake got shallower? Leonard
Susskind: Yeah.
Well, okay.
I took-- so, let's be very specific now.
I kept the lake absolutely uniform height and let's also
suppose that the density of water-- water is an
incompressible fluid.
It can't be squeezed.
It can't be stretched.
Then the velocity vector would be the right think to think
about there.
You could have-- no, you're right.
You could have a velocity vector having a divergence because
the water is-- not because water is flowing in but because
it's thinning out.
Yeah, that's possible.
But let's keep it simple.
And you can have-- the idea of a divergence makes sense in
three dimensions just as much as two dimensions.
You just have to imagine that all of space is filled with
water and there are some hidden pipes coming in, depositing
water in different places so that it's spreading out away
from points in three dimensional space.
Three dimensional space, this is the definition for the
If this were the velocity vector at every point you would
calculate this quantity and that would tell you how much new
water is coming in at each new point in space.
So, that's the divergence.
Now, there's a theorem which the hint of the theorem was
just given by Michael there.
It's called Gauss's theorem and it says something very
intuitively obvious.
You take a surface, any surface.
Take any surface or any curve in two dimensions and now
suppose there's a vector field-- vector field point.
Think of it as the flow of water.
And now let's take the total amount of water that's flowing
out of the surface.
Obviously there's some water flowing out over here and of
course we wanna subtract the water that's flowing in.
Let's calculate the total amount of water that's flowing out
of the surface.
That's an integral out of the surface.
Why is it an integral in because we have to add up the flows
of water outward when the water is coming inward that's just
negative flow, negative outward flow.
We add up the total outward flow by breaking up the surface
into little pieces and asking how much flow is coming out
from each little piece here? How much water is passing out
through the surface? If the water is incompressible,
incompressible means its density is fixed and furthermore,
the depth of the water is being kept fixed.
There's only one way that water can come out of the surface
and that's if it's being pumped in, if there's a divergence.
The divergence could be over here, could be over here, could
be over here, could be over here.
In fact, anywheres there is a divergence will cause an
effect in which water will flow out of this region here.
So, there's a connection.
There's a connection between what's going on on the boundary
of this region, how much water is flowing throughout
boundary on one hand, and what the divergence is on the
There's a connection between the two and that connection is
called Gauss's theorem.
What it says is that the integral of the divergence in the
interior, that's the total amount of flow coming in from
outside, from underneath the bottom of the lake, the total
integrated-- now, by integrated, I mean in the sense of an
The integrated amount of flow in, that's the integral of the
The integral over the interior in the three dimensional case
it would be integral DX, DY, DZ over the interior of this
region of the divergence of A-- if you like to think of A as
the velocity field that's fine-- is equal to the total
amount of flow that's going out through the boundary.
Now, how do we write that? The total amount of flow that's
flowing outward through the boundary we break up-- let's
take the three dimensional case.
We break up the boundary into little cells.
Each little cell is a little area.
Let's call each one of those little areas D sigma.
D sigma, sigma stands for surface area.
Sigma is the Greek letter.
Sigma stands for surface area.
This three dimensional integral over the interior here is
equal to a two dimensional integral, the sigma over the
surface and it is just the component of A perpendicular to
the surface.
It's called A perpendicular to the surface D sigma.
A perpendicular to the surface is the amount of flow that's
coming out of each one of these little boxes.
Notice, incidentally, if there's a flow along the surface it
doesn't give rise to any fluid coming out.
It's only the flow perpendicular to the surface, the
component of the flow perpendicular to the surface which
carries fluid from the inside to the outside.
So, we integrate the perpendicular component of the employee
over the surface, that's still the sigma here.
That gives us the total amount of fluid coming out per unit
time, for example and that has to be equal to the amount of
fluid that's being generated in the interior by the
This is Gauss's theorem.
The relationship between the integral of the divergence and
the interior of some region and integral over the boundary
where it's measuring the flux-- the amount of stuff that's
coming out from the boundary.
Fundamental theorem.
And let's see what it says now.
Any questions about Gauss's theorem here? You'll see how it
I'll show you how it works.
>> Now, you mentioned that the water is compressible.
Is that different from what we were given with the
compressible fluid.
Leonard Susskind: Yeah.
You could-- if you had a compressible fluid you would
discover that the fluid out in the boundary here is all
moving inwards in every direction without any new fluid
being formed.
In fact, what's happening is the fluid's getting squeezed.
But if the fluid can't squeeze, if you can not compress it,
then the only way that fluid could be flowing in is if it's
being removed somehow from the center.
If it's being removed by invisible pipes that are carrying
it off.
>> So that means the divergence in the case of water would
be zero would be integrated over a volume? Leonard Susskind:
If there was no water coming in it would be zero.
If there was a source of the water-- divergence is the same
as its source.
Source of water is-- source of new water coming in from
elsewhere is. . .
So, with the example of the two dimensional lake, the source
is water flowing in from underneath, the sink, which is the
negative of a source, is the water flowing and in the two
dimensional example this wouldn't be a two dimensional
surface integral.
It would be the integral in here equal to a one dimensional
surface equal to the surface coming out.
All right.
Let me show you how you use this.
Let me show you how you use this and what it has the do with
what we've said up 'til now about gravity.
I think-- I hope we'll have time.
Let's imagine that we have a source it could be water but
let's take three dimensional case, there's a divergence of a
vector field, let's say A.
There's a divergence of a vector field, dell dot A, and it's
concentrated in some region of space.
It's a little sphere in some region of space that has
spherical symmetry.
In other words, it doesn't mean that the divergence is
uniform over here but it means that it has the symmetry of a
Everything is symmetrical with respect to rotations.
Let's suppose that there's a divergence of the fluid.
Okay? Now, let's take-- and it's restricted completely to be
within here.
It could be strong near the center and weak near the outside
or it could be weak near the center and strong near the
outside but a certain total amount of fluid or a certain
total divergence, an integrated divergence is occurring with
nice spherical shape.
Let's see if we can use that to figure out what the A field
That's dell dot A in here and now let's see can we figure
out what the field is elsewhere outside of here? So, what we
do is we draw a surface around there.
We draw a surface around there and now we're going to use
Gauss's theorem.
First of all, let's look at the left side.
The left side has the integral of the divergence of the
vector field.
All right.
The vector field or the divergence is completely restricted
to some finite sphere in here.
What is-- incidentally, for the flow case, for the fluid
flow case, what would be the integral of the divergence?
Does anybody know? It really was the flow of a fluid.
It'll be the total amount of fluid that was flowing in per
unit time.
It would be the flow per unit time that's coming through the
But whatever it is, it doesn't depend on the radius of the
sphere as long as the sphere, this outer sphere here, is
bigger than this region.
Why? Because the integral over the divergence of A is
entirely concentrated in this region here and there's zero
divergence on the outside.
So, first of all, the left-hand side is independent on the
radius of this outer sphere, as long as the radius of the
outer sphere is bigger than this concentration of divergence
So, it's a number.
Although it's a number.
Let's call that number M.
No, no.
Not M.
That's the left-hand side.
And it doesn't depend on the radius.
On the other hand, what is the right side? Well, there's a
flow going out and if everything is nice and spherically
symmetric then the flow is going to go radially outward.
It's going to be a pure, radially outward directed flow if
the flow is spherically symmetric.
Radially outward directed flow means that the flow is
perpendicular to the surface of the sphere.
So, the perpendicular component of A is just the magnitude
of A.
That's it.
It's just the magnitude of A and it's the same everywheres
on the sphere.
Why is it the same? Because everything has spherical
Now, in spherical symmetry, the A that appears here is
constant over this whole sphere.
So, this integral is nothing but the magnitude of A times
the area of the total sphere.
All right? If I take an integral over a surface, a spherical
surface like this, on something that doesn't depend on where
I am in the sphere, then it's just you can take this on the
outside, the magnitude of the field and the integral D sigma
is just the total surface area of the sphere.
What's the total surface area of the sphere? >> Four thirds
pi R.
Leonard Susskind: No third.
Just four pi.
Four pi R squared.
Oh, yeah.
Four pi R squared times the magnitude of the field is equal
to Q.
So, look what we have.
We have that the magnitude of the field is equal to the
total integrated divergence divided by four pi.
Four pi is just a number, times R squared.
Does that look familiar? It's a vector field.
It's pointed radially outward.
Well, it's pointed radially outward if the divergence is
If the divergence is positive, it's pointed radially outward
and its magnitude is one of R squared.
It's exactly the gravitational field after a point particle
of the center.
>> It's the magnitude of A.
Leonard Susskind: Yeah.
That's why we have to put a direction in here.
You know what this R is? It's a unit vector pointing in the
radial direction.
It's a vector of unit length pointed in the radial
Right? So, it's quite clear from the picture that the A
field is pointing radially outward.
That's what this says here.
In any case, the magnitude of the field that points radially
outward, it has magnitude Q, and it falls off like one over
R squared.
Exactly like the Newtonian field of a point mass.
So, a point mass can be thought of as a concentrated
divergence of the gravitational field right at the center.
A point mass.
A literal point mass can be thought of as a concentrated-- a
concentrated divergence of the gravitational field.
Concentrated in some very little small volume.
Think of it, if you like, you can think of it as the
gravitational field, the flow field, the velocity field of a
fluid that's spreading out.
Oh, incidentally, of course, I've got the sign wrong here.
The real gravitational acceleration points inward which is
an indication that this divergence is negative.
The divergence is more like a convergence sucking fluid in.
So the Newtonian gravitational field is isomorphic, is
mathematically equivalent, or mathematically similar, to a
flow field to a flow of water or whatever other fluid where
it's all being sucked out from a single point and, as you
can see, the velocity field itself or in this case the
field, the gravitational field, the velocity field would go,
like, one over R squared.
That's a useful analogy.
That is not the say that space is a flow or anything.
It's a mathematical analogy that's useful to understand the
one over R squared force law that it is mathematically
similar to a field of velocity flow from the flow that's
being generated right at the center of a point.
That's a useful observation.
But notice something else.
Supposing now, instead of having the flow concentrated at
the center here, supposing the flow is concentrated over a
sphere that is bigger but the same total amount of flow.
It would not change the answer.
As long as the total amount of flow is fixed, the way that
it flows out through here is also fixed.
This is Newton's theorem.
Newton's theorem in the gravitational context says that the
gravitational field of an object, outside the object is
independent of whether the object is a point mass at the
center or whether it's a spread out mass, or whether it's a
spread out mass this big, as long as you're outside the
object and as long as the object is spherically symmetric,
in other words, as long as the object is shaped like a spear
and you're outside of it, outside of it, outside of where
the mass distribution is, then the gravitational field of it
doesn't depend on whether it's a point, it's a spread out
object, whether it's denser at the center and less dense on
the outside, less dense at the center and more dense on the
All it depends on is the total amount of mass.
The total amount of mass is like the total am of flow coming
into the-- that theorem is very fundamental and important to
thinking about gravity.
For example, supposing we are interested in the motion of an
object near the surface of the earth but not so near that we
can make the flat space approximation.
Let's say at a distance two, three, or one and a half times
the radius of the earth.
Well, that object is attracted by this point, it's attracted
by this point, it's attracted by that point.
It's close to this point, it's far from this point.
It sounds like a hellish problem to figure out what the
gravitational effect on this point is.
But no.
This tells you the gravitational field is exactly the same
as if the same total mass was concentrated right at the
Okay? That's Newton's theorem.
It's a marvelous theorem.
It's a great piece of luck for him because without it he
couldn't have solved his equations.
He knew.
He had an argument, it may have been essentially this
I'm not sure what argument he made.
But he knew that with the one over R squared force law and
only the one over R squared force law wouldn't have been
true if it'd be R cubed, R to the fourth, over R to the
With the one over R squared force law, a spherical
distribution of mass behaves exactly as if all the mass was
concentrated right at the center as long as you're outside
the mass.
So, that's what made it possible for Newton to easily solve
his own equations.
That every object, as long as it's spherical in shape,
behaves if it were a point mass.
>> So, if you're down in a mine shaft that doesn't hold?
Leonard Susskind: That's right.
If you're down in a mine shaft it doesn't hold.
But, that doesn't mean that you can't figure out what's
going on.
You can figure out what's going on.
I don't think we'll do it tonight.
It's a little too late.
But yes, we can work out what would happen in a mine shaft.
But that's right.
It doesn't hold in a mine shaft.
For example, supposing you dig a mine shaft right down
through the center of the earth and now you get very close
to the center of the earth.
How much force do you expect to be pulling toward the
center? Not much.
Certainly much less than if all the mass were concentrated
right at the same theory.
You've got the-- it's not even obvious which way the force
is but it's toward the center.
But it's very small.
You displace away from the earth a little bit.
There's a tiny, tiny force.
Much, much less than as if all the mass were squashed
towards the center.
So, right.
It doesn't work for that case.
Another interesting case is supposing you have a shell of
To have a shell of material, think about a shell of source,
fluid flowing in.
Fluid is flowing in from the outside onto this blackboard
and all the little pipes are arranged on a circle like this.
What does the fluid flow look like in different places?
Well, the answer is, on the outside it looks exactly the
same as if everything were concentrated on a point.
But what about in the interior? What would you guess?
Everything is just flowing out away from here and there's no
flow in here at all.
How can there be? Which direction would it be? And so there'
s no flow in here.
>> Wouldn't you have the distance argument? Like, if you're
closer to the surface of the inner shell-- Leonard Susskind:
>> Wouldn't that be more force towards that? No.
See, you use Gauss's theorem.
Let's use Gauss's theorem.
Gauss's theorem says okay, let's take a shell.
The integrated field coming out of that shell is equal to
the integrated divergency.
But there is no divergency here.
So, the net integrated field coming out is zero.
No field on the interior of the shell.
Field on the exterior of the shell.
So, the consequence is that if you made a spherical shell of
material like that, the interior would be absolutely
identical like it would be if there was no gravitational
material there at all.
On the other hand, on the outside, you would have a field
which would be absolutely identical to what happens at the
Now, there is an analogue of this in the general theory of
We'll get to it.
Basically what it says is the field of anything, as long as
it's spherically symmetric on the outside, looks like
identical to the field of a black hole.
I think we're finished for tonight.
Go over divergence and Gauss's theorem.
Gauss's theorem is central.
There would be no gravity without Gauss's theorem.
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